Frequency Response

Given the transfer function $H(z)$, the frequency response is obtained by evaluating it on the unit circle in the complex plane, i.e., by setting $z=e^{j\omega T}$, where $T$ is the sampling interval in seconds, and $\omega$ is radian frequency:^4.3

$$H(e^{j\omega T}) \isdef \frac{1 + g_1 e^{-jM_1\omega T}}{1 + g_2 ... ...ga T}}, \qquad -\pi < \omega T < \pi \qquad\hbox{(Frequency Response)} \protect$$ (4.4)

In the special case $g_1=g_2=1$, we obtain

$$\begin{eqnarray*} H(e^{j\omega T}) &=& \frac{1 + e^{-jM_1\omega T}}{1 + e^{-jM_2... ...{\cos\left(M_1\omega T/2\right)}{\cos\left(M_2\omega T/2\right)} \end{eqnarray*}$$

When $M_1\neq M_2$, the frequency response is a ratio of cosines in $\omega$ times a linear phase term $e^{j(M_2-M_1)\omega T/2}$ (which corresponds to a pure delay of $M_1-M_2$ samples). This special case gives insight into the behavior of the filter as its coefficients $g_1$ and $g_2$ approach 1.

When $M_1=M_2\isdef M$, the filter degenerates to $H(z)=1$ which corresponds to $y(n)=x(n)$; in this case, the delayed input and output signals cancel each other out. As a check, let's verify this in the time domain:

$$\begin{eqnarray*} y(n) &=& x(n) + x(n-M) - y(n-M)\\ &=& x(n) + x(n-M) - [x(n-M... ...) - y(n-3M)]\\ &=& x(n) + y(n-3M)\\ &=& \cdots\\ &=& x(n). \end{eqnarray*}$$